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log(1 + exp(x)) accurately.log1pexp(x)log1p does
not handle these.log(1 + exp(x)).log(1 + exp(x)) accurately.
An adjustment is made when x is positive and large in value.log1p,
exp.x <- c(10, 50, 100, 200, 400, 500, 800, 1000, 1e4, 1e5, 1e20, Inf)
log1pexp(x)
log(1 + exp(x)) # Naive; suffers from overflow
x <- -c(10, 50, 100, 200, 400, 500, 800, 1000, 1e4, 1e5, 1e20, Inf)
log1pexp(x)
log(1 + exp(x)) # Naive; suffers from inaccuracyRun the code above in your browser using DataLab